Problem
Let $P(x)=x^n+64$ be a polynomial. Find the form of the natural number $n$ such that $P(x)=P_1(x)\times P_2(x)$, $\deg P_1(x),\deg P_2(x)\geq1$.
I thought of taking $n \mod 4$. For $n=0\mod 4$ we can use Sophie Germain's identity to show that $P(x)$ is reducible but after this point I'm stuck.
Any ideas?
Here are the the ones I found
\begin{align} x^{12k+0}+64 =\\ (x^{2k}-2x^k+2)(x^{2k}+2x^k+2)(x^{4k}-2x^{3k}+2x^{2k}-4x^k+4)(x^{4k}+2x^{3k}+2x^{2k}+4x^k+4)\\ x^{12k+3}+64 = (x^{4k+1}+4)(x^{8k+2}-4x^{4k+1}+16)\\ x^{12k+4}+64 = (x^{6k+2}-4x^{3k+1}+8)(x^{6k+2}+4x^{3k+1}+8)\\ x^{12k+6}+64 = (x^{4k+2}+4)(x^{8k+4}-4x^{4k+2}+16)\\ x^{12k+8}+64 = (x^{6k+4}-4x^{3k+2}+8)(x^{6k+4}+4x^{3k+2}+8)\\ x^{12k+9}+64 = (x^{4k+3}+4)(x^{8k+6}-4x^{4k+3}+16), \end{align}
and the ones with the imaginary unit are \begin{align} x^{12k+2}+64 = (x^{6k+1}-8i)(x^{6k+1}+8i)\\ x^{12k+10}+64 = (x^{6k+5}-8i)(x^{6k+5}+8i). \end{align}